Question

From the top of a tower of height 40m a ball is projected upwards with a speed of 20m/s at an angle of elevation of 30°.Then the ratio of the total time taken by the ball to hit the ground to the time taken to the ball to come at the same level as the top of the tower?

Answer 1

Answer: The ratio of the time is 1:3.

Explanation:

Given that,

Height h = 40 m

Velocity u = 20 m/s

Angle of elevation $\theta= 30^{\circ}$

We know that,

$\Delta y= 0$

Using equation of motion

$\Delta y = u \sin\theta\ t -\dfrac{1}{2}gt^2$

$0 = 20\times sin 30^{\circ} t-\dfrac{1}{2}\times9.8\ t^2$

$4.9\ t^2-10t=0$

$t = 2.04\ s$

The ball to come at the same level as the top of the tower at this time.

Now, $\Delta y= -40 m$

Using equation of motion

$\Delta y' = u \sin\theta\ t' +\dfrac{1}{2}gt'^2$

$-40 = 20\times sin30^{\circ}t'-\dfrac{1}{2}\times9.8\ t'^2$

$4.9\ t'^2-10t'-40=0$

$t' = 4.05\ s$

The total time when ball to hit the ground

$t'' = t+t'$

$t''= 2.04+4.05$

$t''= 6.09\ s$

The ball will hit at this time.

The ratio of the total time taken is

$\dfrac{t}{t''}= \dfrac{2.04}{6.09}$

$\dfrac{t}{t''}= \dfrac{1}{3}$  approximate

Hence, The ratio of the time is 1:3.

Answer 2

Your answer is 2:1

Explanation:

Hope it helps ↓